- To participate in the 911Metallurgist Forums, be sure to JOIN & LOGIN
- Use Add New Topic to ask a New Question/Discussion about Crushing.
- OR Select a Topic that Interests you.
- Use Add Reply = to Reply/Participate in a Topic/Discussion (most frequent).
Using Add Reply allows you to Attach Images or PDF files and provide a more complete input. - Use Add Comment = to comment on someone else’s Reply in an already active Topic/Discussion.
Retention time for a nickel ore trommel (4 replies and 2 comments)
Depends on the feed size distribution and the trommel deck openings. Also wether a single deck or multi-deck (and yes they exist). Also need to figure out trommel length and diameter (surface area) and speed of rotation. But in general you are talking underr 2 mintues (often under a minute).
I know Wikiwand isn't a very truthfull website to take info, but is the only one where I found a formula for the residence or retention time on a trommel or rotatory screen. For the formula, the first calculation is α, which is the departure angle of the feed, being this one as:
α= cos-1((r*(Wt)^2)/(g*cosβ))
Where r is the drum radius, Wt is the rotational velocity in radians per second, g is gravity and β is the angle of inclination of the drum. Then, to calculate the retention time tr, it says that:
tr=(L*(360-4α+(229.2*cos α*sin α))) / (48*n*r*tanβ*cos α*(sin α)^2)
The info I used for this is:
r = 1.75 m , L = 7 m , Wt = 1.2 rad/s , g = 9.8 m/s , n = 11.5 rpm , β = 0.001°
Using this formula, where there is no feed size variable, I have that the retention time using 11.5 rpm (the only variable we can change during the mineral processing) is about 40 seconds, and changing the velocity to 7 rpm, change the time to almost 3 minutes.
If you are trying to be truly theoretical, those equations are reasonable. But you really do need to take the feed size distribution into account as well as the trommel opening size. Those equations are basically for a tumbling drum such as a dryer or attrition machine.
So I have to attach the feed size distribution and the trommel opening to the equation, in this case the maximum feed size is about 190mm but for the trommel opening size, will it be like the area of which the feed is going into the trommel? Or it is about the size of the first set of openings in which the mineral will be concentrated? I'm sorry if I can't explain it properly.
A trommel is a rotating screen, its purpose to separate oversized from under sized material. The trommel opening I mentioned is the size of holes in the trommel deck. Then you need to consider how much material in the feed is coarser than that size , how much is near to that size, and how much is under that size. Then with a larger enough diameter to break up clumps and easily broken pieces.
In general trommel sizing is based (like any screening/classifying device) on the amount of material through the deck. If properly sized, 75 to 80 % of the under sized material will discharge in the first 1/2 of the trommel, with the remaining half for breaking up of the easily broken pieces.
Hello everyone,
I have been looking for a formula to the retention time on a mineral rotatory trommel but I'm unsure about one I found in the internet. I want to know if there is a formula for that so I can calculate the average or theorical retention time for the trommel that is coming to the mine I work with before we have to do the hot testing of the equipment.
In my team we have run a few test on a minor scale trommel on site but these one has a closed processing of the minerals, different to the new one that has a continous feed and discharge mineral screening and processing.
I really appreciate your help.